Question

show that the square of any positive odd integer is of the form 4q + 1 where q is any integer​

Answer 1

Answer:

Step-by-step explanation:

Let positive integer a = 4m + r ,

By division algorithm we know here 0 ≤ r < 4 ,  

When r = 0

a = 4m  

Squaring on both sides,

a^2 = ( 4m )^2

a^2 = 4 ( 4m^​2)

a^2 = 4 q   (  q = 4m^2  ) (Even)

When r = 1

a = 4m + 1

squaring on both sides,

a^2 = ( 4m + 1)^2

a^2 = 16m^2 +  8m   + 1

a^2 = 4 ( 4m^2 + 2m ) + 1  

a^2 = 4q + 1   (  q = 4m^2 + 2m  ) (Odd )

When r = 2

a = 4m + 2  

Squaring on both  sides,

a^2 = ​( 4m + 2 )^2

a^2 = 16m^2 +  16m  + 4

a^2 = 4 ( 4m^2 + 4m + 1 )

a^2 = 4q       ( q = ​ 4m^2 + 4m + 1  ) (Even)

When r = 3  

a = 4m + 3

Squaring on both sides,

a^2 = ​( 4m + 3)^2

a^2 = 16m^2 +  24m  + 9

a^2 = 16m^2 + 24m ​ + 8 + 1

a^2 = 4 ( 4m^2 + 6m + 2) + 1

a^2 = 4q + 1     ( q = 4m^2 + 6m + 2   )  (Odd )

So,Square of any positive odd integer is in form of  4q + 1 , where q is any integer.